## Physics (10th Edition)

Published by Wiley

# Chapter 4 - Forces and Newton's Laws of Motion - Problems: 53

#### Answer

1.00 $\times$ $10^{2}$ N, 53.1$^{\circ}$ south of east

#### Work Step by Step

$F_{x}$ = 60 N east $F_{y}$ = 80 N south F = $\sqrt 60^{2}$ + $\sqrt 80^{2}$ = 100 N, $tan^{-1}$($\frac{80}{60}$) $\approx$ 100 N, 53.1$^{\circ}$ south of east

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